Download - EE4031 5 Transient Stability

The HK Polytechnic University Transient Stability Analysis

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1 Power System Stability• A large power system consists of a number of synchronous machines operating

in synchronism – synchronous generators located at great and small distances

apart and of varied ratings and characteristics have to continuously operate at

the same frequency (electrical speed = pole pairs × mechanical speed).

• It is necessary that they should maintain synchronism despite the ever present

disturbances induced by nature and by man – load changes, switching actions,

faults and so on.

• When the system is subject to some form of disturbance, there is a tendency for

the system to develop forces to bring it to a normal or stable condition. The

ability of a system to reach a normal or stable condition after being disturbed is

call stability.

• Synchronous stability may be divided into two main categories depending upon

the magnitude of the disturbance – steady-state, dynamic and transient stability.

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• The steady-state stability is the ability of a system to bring it to a stable

condition after a small disturbance. The study of steady-state stability is

basically concerned with the effect of gradual infinitesimal power changes and

the dynamics of rotating machines will be excluded from the studies.

• However in practice the dynamics of rotating machines will be effected even the

increase in load is gradual. Dynamic stability is an extension of steady-state

stability where the dynamic effects of synchronous machines and automatic

control devices such as governors and voltage regulators are included. The

dynamic stability is concerned with small disturbances lasting for a long time

and in dynamic stability studies non-linearities are neglected.

• The transient stability is the ability of a system to bring it to a stable condition

after a large disturbance. Transient stability is concerned with sudden and large

changes in the network conditions. The large disturbances can occur due to

sudden changes in application or removal of loads, line switching operations,

line faults, or loss of excitation.

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• A mechanical analog of power system transient stability is given in the figure

below, in which a number of masses representing synchronous machines are

interconnected by a network of elastic strings representing transmission lines.

When one of the strings is cut, representing the loss of a transmission line, the

masses undergo transient oscillations and the forces on the strings fluctuate.

The system will then either settle down to a new steady-state operating point or

additional strings will break, resulting in an even weaker network and eventual

system collapse.

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2 Stability Limits

• The stability limit is the maximum power that can be transferred in a network

between sources and loads without loss of synchronism.

• The steady-state stability limit is the maximum power that can be transferred

without the system becoming unstable under steady-state conditions.

• The dynamic stability limit is the maximum power that can be transferred without

the system becoming unstable under small disturbances lasting for a long time.

• The transient stability limit is the maximum power that can be transferred without

the system becoming unstable when a sudden or large disturbance occurs.

• In general, the transient stability limit is the lowest whilst the steady-state limit is

the largest.

i.e. transient stability limit < dynamic stability limit < steady-state stability limit

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3 Synchronism and Steady-State Stability

3.1 Two Finite Machines

• First consider case of two identical unloaded machines that are subsequently

disturbed.

E1

X X

E2

E2 E2E1

E1

Ic

E - E1 2

I =c2jX

E - E1 2

Ic

• Say some disturbance occurs that causes generator 1 to accelerate with

respect to generator 2. Observed that power comes out of the faster machine

terminals and passes into the slower machine terminals. The disturbance is

corrected, opposed and (hopefully) overcome.

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• Suppose the machines were initially loaded and equally sharing a load. Apart

from losses, both machines would have a mechanical input equal to the

electrical output.

E1

X X

E2

E2 E2E1

E1

Ic

E - E1 2

IcI1 I2

I = I1 2

I1

I2

• Once again say the same disturbance causing generator 1 to accelerate with

respect to generator 2 were to occur. The effects of this are shown in the above

figure. As can be seen, the main effect is to increase the electrical burden on

the faster machine and to reduce it on the slower machine. The resulting power

unbalance with respect to mechanical input tends to correct the disturbance.

The power flow between the generators is called the synchronous power flow.

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3.2 Infinite Bus

• In a power system, normally more than two generators operate in parallel. The

machines may be located at different places. A group of machines located at

one place may be treated as a single large machine. Also, the machines not

connected to the same bus but separated by lines of low reactance, may be

grouped into one large machine.

• The operation of one machine connected in parallel with such a large system

comprising many other machines is of great interest. The capacity of the system

is so large that its voltage and frequency may be taken constant. The

connection or disconnection of a single small machine on such a system would

not affect the magnitude and phase of the voltage and frequency.

• Such a system of constant voltage and constant frequency regardless of the

load is called infinite busbar system or simply infinite bus. Physically it is not

possible to have a perfect infinite bus.

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3.3 Single Machine on Infinite Bus

The stability problem can be studied with an analysis of the behaviour of a

synchronous generator through a line to an infinite busbar as shown below.

jX

V 0oE δ δ

VEInfinite Bus

where V = phase voltage of the infinite bus

E = excitation phase voltage of the generator

X = total synchronous reactance from the source to infinite bus

δ = load angle, i.e. phase angle between V and E

and the power transfer P from the generator to the infinite bus is given by

P =EV

Xsin δ

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Maximum power Pmax will be transferred when δ = 90o, i.e.

Pmax =EV

X

and the power transfer can be rewritten as

P = Pmax sin δ (3.3.1)

Equation (3.3.1) represents the steady-state stability limit of the power system. It is

clean from the equation that steady-state stability limit Pmax can be increased by

1. Increasing system voltages E or V : – by increasing the excitation

2. Decreasing system reactance X by : – use of double-circuit line

– use of bundled conductors

– series compensation of the reactance

– use of machines of low impedances

In fact, increasing steady-state stability limit by decreasing reactance is economical

and also effective.

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The graphical representation of power received P and the load angle δ is called the

power-angle diagram as shown below.

1. Magnitude of P (output) depends on

magnitude of δ.

2. There is a maximum limit to the power

that can be extracted from a generator for

given E, V and X .

3. The sign of the output power depends on

the sign of δ (Generating/Motoring).

4. There is a stable range of operation from

δ = -90o to δ = 90o and stable operation

is not possible outside this range.

Pmax

-Pmax

90 180-90-180

Power angle curve

Generator

Motor

P

δ

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3.4 Synchronizing Power (and Torque) Coefficient

The system is stable if and only if for an increase in rotor angle δ (load) the

transmitted power also increases, i.e. the dPdδ

should be positive.

The rate dPdδ

is called the synchronizing power coefficient and is taken as the

measure of the stability of a system, i.e.

synchronizing power coefficient, ps =dP

dδ= Pmax cos δ

Hence, the steady-state synchronous stability criterion for a simple system is

ps > 0, i.e. the synchronizing power coefficient is positive. The steady-state

stability limit is reached when ps = 0 and if ps < 0, then system is unstable.

If ω0 is the synchronous speed, then

synchronizing torque coefficient, ts =ps

ω0=

Pmax

ω0cos δ

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4 Synchronous Machine Models

4.1 The Voltage Behind Synchronous Reactance Model

• The previous model in which X = Xs the synchronous reactance is true only

for slowly changing conditions. Since resistance is neglected, there should be

no significant transmission line length prior to connection to a large busbar

system. Since saliency is neglected, the machine should be of cylindrical pole

construction.

• Most of these assumptions are rarely true. The most unsatisfactory one is

regarding slow changes. This assumption is not acceptable for transient and

dynamic stability studies.

4.2 Connection via a Transmission LinejX

V 0oE δ

Infinite BusjXg l R l

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Let Z = R + jX be the total impedance where

R = Rl

X = Xg + Xl

and Z = |Z|/θ = |Z|/90o − α and α is small.

The power transferred can be obtained as

P =EV

|Z|sin(δ + α) −

V 2

|Z|sin α

=EV

|Z|sin(δ + α) −

V 2

|Z|R where sin α =

R

|Z|

If the line is long enough, it may be necessary to incorporate a T or π model for the

line. jX

V 0oE δg

V 0oE δy12

y22y11

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By well known methods of network reduction the circuit can be reduced to a single π

equivalent as shown in the above figure with all circuit quantities being shown in

admittance form.

Now, since

I1

I2

=

Y11 Y12

Y21 Y22

E/δ

V /0o

where Y11 = y12 + y11 ; Y12 = Y21 = −y12 and Y22 = y12 + y22

Hence, P = EV |Y12| sin(δ + γ) + E2G12

where Y12 = |Y12|/90o − γ and G12 = Re(Y12)

4.3 The Voltage Behind Transient Reactance Model

None of the previous models are really satisfactory because they overlook the fact

that during the first one or two seconds after a disturbance the flux linkages of the

field circuit cannot be easily altered. For a much shorter period the same is true of

the flux linkages of the damper circuits and solid rotor iron also.

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Hence for transient and dynamic stability studies a more sophisticated machine

model is required.

The equivalent circuit and phasor

diagram for a cylindrical pole rotor

machine are shown.

Since X ′

d < Xd, the transient

power-angle curve is much

higher than the steady-state one.

P =E′V

X ′

d

sin δ

V

E’

X’ R

I

δ

o/V

E

IR IX’

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5 Transient Stability• The steady-state stability limit does not provide a useful indication because the

actual disturbance that occur in power systems can be very large (faults, loss of

generators, line switching, major load changes, etc.) and the system should be

operated to be stable through these events.

• The transient stability is the ability of a system to maintain synchronous

operation and to reach a stable state (or the one close to it) after a large

disturbance.

• Different utiilies may specify different criterion disturbances. For example the

criterion disturbance could be a 3-phase short circuit which is cleared by circuit

breaker operation in a time corresponding to the system fault clearing time.

• The severity of a disturbance and the stability limit depend on

1. system and generator conditions (voltages, impedances, etc.)

2. the point of fault (location) and its duration

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3. control loops such as AVR’s and governors

4. mechanical factors such as boiler and hydraulic time constants

5. load characteristics

• Under normal operations, the relative position of the rotor axis and stator

magnetic field axis is fixed. The angle between the two is known as the load

or rotor angle δ and depends upon the loading of the machine.

• The problem of stability revolves around the determination of whether or not

the torque and rotor angle δ will stabilize after a sudden disturbance. In case

δ continues to increase after a disturbance the machine will lose synchronism.

• In order to determine the angular displacement between the generating units of

a power system following a major disturbance, it is necessary to solve a set of

differential equations, called the swing equations, describing the motion of the

rotors.

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5.1 Swing Equation

In a synchronous generator the input is the mechanical or shaft torque Tm and the

output is the electrical torque Te. If the losses are neglected the difference between

the mechanical torque and the electrical torque is equal to the accelerating or

decelerating torque Ta, i.e.

Ta = Tm − Te

If J is the moment of inertia of the rotor and α is the angular acceleration,

the motion of the rotor can be expressed as:

Ta = Tm − Te = Jα

Let θ be the angular position of the rotor at any instant t,

δ be the corresponding angular displacement of the rotor and

ω, ωs be the angular velocity and synchronous velocity respectively.

θ = ωst + δ ω =dθ

dt= ωs +

dδ

dtα =

d2θ

dt2=

d2δ

dt2

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Hence, the rotor motion equation can be written as:

Tmω − Teω = Jωα

or Pm − Pe = Mα = Md2δ

dt2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5.1.1)

where Pm = Tmω the mechanical power

Pe = Teω the electrical power

M = Jω the angular momentum

Equation (5.1.1) is known as the swing equation and describes the behaviour

of a synchronous machine during transients.

As the angular momentum M is proportional to the angular velocity, it is not

constant but varies somewhat during the swings due to variation in ω. In practice,

the change in ω from the normal system angular velocity ωs is not much during the

swing and it is commonly assumed that M is constant and is equal to Jωs. This

value of M is known as the inertia constant of the machine.

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Another important constant denoted by H and is defined as the ratio of the kinetic

energy at rated speed to the rated apparent power of the machine, i.e.

H =stored energy in megajoules

rating in MVAMWs/MVA or MJ/MVA

H is also sometimes called as inertia constant. A relation between M and H is

derived as follows:

GH = energy stored = 12Mω = 1

2M · 2πf or M =GH

πfwhere G is the rating of the machine in MVA and f is the system frequency.

Note M depends upon the size of the machine as well as on its type whereas

H does not vary widely with size and has a characteristic value or set of values

for each class of machines.

Substituting M in Equation (5.1.1):GH

πf0

d2δ

dt2= Pm − Pe

or in per unitH

πf0

d2δ

dt2= Pm − Pe . . . . . . . . . . (5.1.2)

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where f0 is the rated (synchronous) frequency.

Hence for a N multi-machine system, the swing equations are

Hi

πf0

d2δ

dt2= Pmi − Pei for i = 1, 2, ... , N . . . . . . . . . . . . . . (5.1.3)

Damping has been ignored in the derivation of the above relation. The solution of

Equation (5.1.3) provides the values of δ with respect to time t which when plotted

gives the swing curve. Thus, for a multi-machine system, there will be a swing curve

for each generator and the transient stability of the system is assessed by observing

the angular difference between each pair of generators.

If the difference is found to be decreasing after all switching operations are over, the

system will be generally stable and the oscillation will be successively damped out.

However, it is possible that the units may go out of synchronism after withstanding

the first swing successfully – multi-swing stability.

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Swing curves provide information regarding stability and show any tendency of δ to

oscillate and/or increase beyond the point of return.

Unstable

Stable

δ

t

Unstable

Stable

δ

t

First-swing stability Multi-swing stability

Swing curves are also useful in determining the adequacy of relay protection on

power system with regard to the clearing of faults before one or more machines

become unstable and fall out of synchronism. The critical clearing time is found to

specify the correct speed of the circuit breaker.

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Example 5.1.1

A 50Hz, 20 MVA, 13.2kV generator has an inertia constant H = 9 MWs/MVA.

Determine :

a) the K.E. stored in the rotor at synchronous speed.

b) the acceleration if the net mechanical input power is 18 MW and the electrical

power developed is 15 MW.

c) the change in rotor angle for a period of 15 cycles with rotor acceleration being

constant.

Solution:

a) K.E. stored in the rotor in megajoules

= H× machine rating in MVA

= 9 × 20 = 180MJ

b) The accelerating power, Pa = Pm − Pe = 18 − 15 = 3MW

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Now the acceleration is

d2δ

dt2=

Pa

Mwhere M =

GH

πfo

=20 × 9

π × 50= 1.146 MJsec/rad

⇒d2δ

dt2=

3

1.146= 2.618 rad/sec2

c) Again using the swing equation,

dω

dt=

d2δ

dt2= 2.618 ⇒ ω = 2.618 t rad/sec

and

dδ

dt= ω = 2.618 t ⇒ δ = 1.309 t2 rad

Period of 15 cycles =15

50= 0.3 sec

Hence, δ = 1.309 × 0.32 = 0.11781 rad = 6.75 deg

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5.2 Transient Stability Assessment

For a two machine system under the assumption of constant input, no damping and

constant voltage behind transient reactance, the machines either fall out of step in

the first swing or never. Under this condition the two machines are said to be

running at standstill with respect to each other.

There is a graphical method of determining whether the two machines are running

at standstill with respect to each other or not. This method is known as Equal Area

Criterion for stability. The use of this method eliminates partially or wholly the

calculation of swing curves which thus saves a considerable amount of work.

The Equal Area Criterion method is applicable to any two machine systems which

satisfies the above assumption. It is not applicable to a multi-machine system

directly or if the critical clearing time is required to be determined. In such case,

step-by-step (or point-by-point) method may be used for numerical solution of swing

equations.

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5.3 Equal Area Criterion

The equal area criterion may be used to assess the transient stability of two

machine system or one machine connected to an infinite bus without actually

solving the swing equation. It is derived using the swing equation given as:

Md2δ

dt2= Pa = Pm − Pe

Multiplying both the sides of the equation by 2dδ

dtand integrating w.r.t. time:

∫

2Md2δ

dt2

dδ

dtdt =

∫

2(Pm − Pe)dδ

dtdt

or M

(

dδ

dt

)2

= 2

∫

δ

δ0

(Pm − Pe)dδ

ordδ

dt=

√

2

M

∫

δ

δ0

(Pm − Pe)dδ + constant

where δ0 is the initial rotor angle before any disturbance occurs.

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Also when t = 0, dδdt

= 0 and δ = δ0, i.e. the constant term in the above equation

is zero. The angle δ will stop changing and the machine will again be operating at

synchronous speed after a disturbance when dδdt

= 0 or when

∫ δ

δ0(Pm − Pe)dδ = 0 =

∫ δ

δ0Padδ

This means that the area under the curve Pa should be zero. This condition is

possible only when Pa is both accelerating and decelerating powers.

For a generator action Ps > Pe for positive area

A1 and Pe > Ps for negative area A2 for stable

operation. Hence the name equal area criterion.

The area A1 represents the kinetic energy stored

by the rotor during acceleration and the area A2

represents the kinetic energy given up by the rotor

to the system and when it is all given up, the

machine has returned to its original speed.

A1

A2Ps

Pe

δδ0

P

+ve

-ve

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Consider the following sequence of events. A generator working initially and stably

at an “ input = output = P0” operating point with torque angle δ0 is subjected to a

sudden increase of input to level P1. The generator should settle down to new

stable operation at P1, δ1.

P

P0

P1

δ0 δ1

X

Yb

a

δ

1. Rotor begins to accelerate

2. Point P1, δ1 reached and there is power

balance but the large rotor mass is overspeed

3. The rotor runs past P1, δ1

4. The output > input therefore deceleration

commences

5a. At some point such as X , speed has fallen again to synchronous speed ωs but

there is no power balance (P1 > Pe)

6a. Speed falls below ωs and the rotor comes back downwards P1, δ1

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7a. The next time P1, δ1 is reached though power balance and exits again with the

rotor mass being underspeed

8a. The rotor runs below P1, δ1 but acceleration commences again

9a. If damping is taken into account, the rotor oscillates about P1, δ1 and settles

down to a stable operation

Alterative scenario – unstable case:

5b. The rotor runs past the steady-state stability limit before speed falls to

synchronous value

6b. The rotor reaches point Y before speed falls to synchronous value

7b. Output falls below input again

8b. The overspeed rotor begins to accelerate again

9b. Synchronism is lost

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In summary, the equal area criterion makes the following assumptions or

approximations:

1. Neglects electro-magnetic, electrical (I2R) and mechanical damping

2. Neglects control loops such as AVR’s and governors

3. Assumes constant power input

None of these are generally true or small. Hence the equal area criterion is

pessimistic – i.e. over-estimates the extent of the expected swing.

5.4 Applications of the Equal Area Criterion

5.4.1 Step increase in input power P

P0

P1

δ0 δ1 δδm

A2

A1

A3

Limit of first swing given by A1 = A2.

Condition for stability given by A1 ≤ A3.

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Now what is the maximum value of the P1 such that the system is critically stable,

i.e. any attempt to increase P1 beyond this value the system becomes unstable ?

Referring to the power-angle curve plot

P1 = Pmax sin δ1 = Pmax sin δm ⇒ δm = (π − δ1)

The value of δ1 with this condition is known as critical torque angle δc, i.e. δc = δ1.

Also the following condition should be satisfied for A1 = A3:

P1(δm − δ0) =

∫ δm

δ0

Pmax sin δdδ

Pmax sin δm(δm − δ0) = Pmax(cos δ0 − cos δm)

Here δm is the only unknown which can be obtained and hence the critical P1 can

be calculated.

It is to be noted that δ0 must lie within the steady-state limit, i.e. δ0 < 90o, but δ1

may be greater than 90o as long as the equal area criterion is satisfied.

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5.4.2 Sustained network fault – Line switching

Consider a system represent by a synchronous generator connected to an infinite

busbar by a double-circuit line. The two circuits are connected in parallel.

Line 1

Line 2

Generator

P

P1

P0

δ0 δ1 δδm

A2

A1

A3

Double LinePAC

Single LinePAC

If one of the circuits is switched out the system may

become unstable in spite of the fact that the load

could be supplied over by the other circuit under

steady-state conditions.

Applying the equal area criterion:

Limit of first swing given by A1 = A2.

Condition for stability given by A1 ≤ A3.

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5.4.3 Network fault subsequently cleared – Faulted line

Suppose that a fault occurs on one of the two parallel circuits connecting to an

infinite busbar system.

Line 1

Line 2

Generator

fault

The faults produces a transient change which may render the system unstable.

However, if the circuit breakers clear the fault in time, it is possible to maintain

stability. The maximum value of time allowed for protective gear to operate without

loss of stability is called the critical clearing time. The torque angle corresponding to

this time is called critical clearing angle.

Let δcl be the value of δ at which the fault is cleared.

Limit of swing is given by A1 = A2 and the condition for stability by A1 ≤ A3.

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Observe the influence of both severity and duration of fault on stability. This also

shows the need for fast protection relaying and circuit breaker clearing of faults. On

important e.h.v. or h.v. systems, 0.5 to 1 cycle operation of protection and 1 to 3

cycle circuit breaker clearing is now possible to achieve.

Let the three power angle curves (PACs) be

represented as

Before the fault Pmax sin δ

During the fault r1Pmax sin δ

After the fault r2Pmax sin δ

P0

δ0 δcl δδm

A2

A1

A3

Pre-faultPAC

Post-faultPAC

FaultedPAC

Pmax

r Pmax2

r Pmax1

For transient stability limit, the two ares A1 = A3

P0(δm − δ0) =

∫ δcl

δ0

r1Pmax sin δdδ +

∫ δm

δcl

r2Pmax sin δdδ

= r1Pmax(cos δ0 − cos δcl) + r2Pmax(cos δcl − cos δm)

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Substituting P0 = Pmax sin δ0 and rearranging the above equation into:

cos δcl =(δm − δ0) sin δ0 − r1 cos δ0 + r2 cos δm

r2 − r1

Also from the curves: P0 = Pmax sin δ0 = r2Pmax sin(π − δm)

or δm = π − sin−1

(

sin δ0

r2

)

Example 5.4.1

Generator fault Infinite System

j0.3j0.15 j0.25 j0.15

j0.151.2pu 1pu

j0.15 j0.15j0.25

P1pu

All impedances are given in ohms. Determine the critical clearing angle for the

generator for a 3-φ fault at the point P when the generator is delivering 1 pu power.

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Solution :

The total transmission reactance

• before the fault : Xpre = 0.3 + 0.552 + 0.15 = 0.725 pu

• during the fault : Xflt = ZA′B′ = ZAZB+ZBZC+ZCZA

ZC

= 0.375×0.35+0.35×0.0545+0.375×0.05450.0545

= 3.133 pu

• after the fault : Xpost = 0.3 + 0.55 + 0.15 = 1.0 pu

i.e. The maximum power output is given as

• before the fault : Pmax = EVXpre

= 1.2×1.00.72 = 1.667 pu

• during the fault : Pmax1= EV

Xflt= 1.2×1.0

3.133 = 0.383 pu

• after the fault : Pmax2= EV

Xpost= 1.2×1.0

1.0 = 1.2 pu

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And the value of r1 =Pmax1

Pmax

=0.383

1.667= 0.23

r2 =Pmax2

Pmax

=1.2

1.667= 0.72

The rotor angle, δ0, when the system is operating normally

Pmax sin δ0 = 1 pu ⇒ δ0 = sin−1 11.667 = 36.9o = 0.643 rad

When the system critically stable, Pmax2sin(π − δm) = Pmax sin δ0

i.e. Pmax2sin δm = 1 pu ⇒ δm = sin−1 1

1.2 = 123.6o = 2.156 rad

Using the critical clearing angle equation


r2 − r1

=(2.156 − 0.643) × 0.6 − 0.23 × 0.8 − 0.72 × 0.553

0.72 − 0.23

⇒ δcl = 48.35o

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5.5 Step-by-Step (Point-by-Point) Method

• The equal area criterion (EAC) method is useful in determining the critical

clearing angle, i.e. the condition when the system will be stable provided the

fault is cleared before the rotor angle exceeds the critical clearing angle.

• However the EAC method provides little information on the critical clearing time

which is important for setting up the operation times of the relay and circuit

breaker such that the total time taken by them would less than the critical

clearing time for stable operation of the system.

• The step-by-step or point-by-point method is the method which can determine

the critical fault clearing time. This method can also be used for the solution of

multimachine system. At present, it still is one of the most practical avaliable

methods of transient stability analysis for power systems with controls.

• Because this method involves the calculations of the rotor angles as time is

incremented, it is also referred to as the time domain simulation method.

KWCn v2.2 19


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• Numerical integration techniques are used to solve the swing equations. Each

swing curve (δ − t) is then plotted to check its tendency to remain stable.

• The step-by-step calculations can be made by hand, the ac network analyser or

the digital computer. The accuracy of the solution depends upon the time

incremnet used in the analysis. As the time interval is decreased, the computed

swing curves approach the ture curve.

• As the step-by-step method calculates the change in the angular position of the

rotor during a short interval of time (milliseconds), the following assumptions are

commonly made during the computational procedure.

1. The accelerating power Pa and the angular acceleration ( d2δdt2

) computed at

the beginning of an interval is assumed to be constant from the middle of the

preceding interval to the middle of the interval under consideration.

2. The angular velecity ω computed at the middle of an interval remains

constant over the interval.

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timestep

nn-1n-2 timestep

nn-1n-2 timestep

nn-1n-2

Assumed

Actual

Pa(n)

Pa(n-1)Pa(n-2)

δ(n)

δ (n-1)

δ

w(n)

wPa

calculation point region assumed constant

• The above assumptions are not strictly correct, since δ is changing continuously

and both Pa and ω depend upon δ. However, they are acceptable if time

increment ∆t is made small.

5.5.1 Simple Step-by-Step Numerical Integation

First choose a small time step ∆t, say 0.01 to 0.05 sec.

KWCn v2.2 20


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The procedure for the n-th interval is outlined below:

1. Evaluate the accelerating power Pa(n−1) = Pm(n−1) − Pe(n−1)

2. From the swing equationd2δ

dt2= α(n) =

1

MPa(n−1)

where α is the acceleration, evaluate α(n).

3. The change in angular velocity for the n-th interval is

∆ω(n) = α(n)∆t

Hence, ω(n) = ω(n−1) + ∆ω(n) = ω(n−1) + α(n)∆t

Here ω is the relative angular velocity and is zero at t = 0.

4. The change in rotor angle for the n-th interval is

∆δ(n) = ω(n)∆t = [ω(n−1) + ∆ω(n)]∆t = ∆δ(n−1) + ∆ω(n)∆t

= ∆δ(n−1) + α(n)(∆t)2

Hence, δ(n) = δ(n−1) + ∆δ(n) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (5.5.1)

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Equation (5.5.1) forms the basis of the numerical solution by step-by-step method.

The accelerating power (based on the powers calculated for last interval) is

calculated at the beginning of each new interval.

In other words, once the value of δ(n) obtained from this equation, it forms one point

on the swing curve and will be used for evaluating the accelerating power Pa(n+1)

and the procedure will be repeated during the subsequent intervals.

Pa(n+1) = Pm(n) − Pe(n)

where Pe(n) =EV

Xsin δ(n)

and Pm(n) = constant for classical modelling

Once the swing curve is plotted and the critical clearing time, corresponding to the

critical clearing angle as obtained by using the equal area criterion, can be

determined and a suitable protection scheme can be designed to avoid the system

from falling out of step.

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Example 5.5.1

Three-phase fault takes place at B as shown in the figure below with all line

impedances being given in Ω. The breakers at A and B operate simultaneously.

The generator has an inertia constant H = 4.0 and it is delivering 1.0 pu power

before the fault takes place. Use a time step of 0.05 sec.

j0.25

j0.3

Generator

faultj0.12

j0.035

1.1pu

Infinite Bus

1puBA

1. Determine the critical fault clearing angle.

2. Determine the critical clearing time.

3. Plot the swing curves both under sustained fault and when the breakers have

operated at the end of 10 cycles, i.e. 0.2 s.

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Solution :

1. Critical clearing angle calculation:

Pre-fault reactance between the buses

Xpre = 0.12 + 0.25//0.3 + 0.035 = 0.29136 Ω

Post-fault reactance between the buses

Xpost = 0.12 + 0.25 + 0.035 = 0.405 Ω

Maximum power delivered :

• when the system is operating normally

Pmax = EVXpre

= 1.1×1.00.29136 = 3.775 pu

• during the fault

Pmax1= 0 pu

• after the breakers operate

Pmax2= EV

Xpost= 1.1×1.0

0.405 = 2.716 pu

⇒ r1 =Pmax1

Pmax= 0 & r2 =

Pmax2

Pmax= 2.716

3.775 = 0.7194

KWCn v2.2 22


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The rotor angle, δ0, when the system is operating normally

Pmax sin δ0 = 1 pu ⇒ δ0 = 15.36o = 0.2649 rad

When the system critically stable, Pmax2sin(π − δm) = Pmax sin δ0

i.e. Pmax2sin δm = 1 pu ⇒ δm = 158.4o = 2.763 rad

Using the critical clearing angle equation


r2 − r1

=(2.763 − 0.2649) × 0.2649 − 0.7194 × 0.9297

0.7194

⇒ δcl = 90.62o

2. Critical clearing time determination:

M =GH

πf=

1.0 × 4

180 × 50=

1

2250sec2/deg

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First interval:

Since the power transmitted is 0 once the fault applied while the shaft power is

1, the average value of the power in the first interval is (case 2 discontinuity):

Pa0=

Pa0−

+ Pa0+

2=

(1.0 − 1.0) + (1.0 − 0.0)

2= 0.5 pu

Or α0 =Pa0

M= 0.5 × 2250 = 1125 deg/sec2

⇒ ∆ω1 = α0∆t = 1125 × 0.05 = 56.25 deg/sec

ω1 = ω0 + ∆ω1 = 0.0 + 56.25 = 56.25 deg/sec

where ω0 is the angular velocity of the rotor with respect to the 50 Hz

reference axis and is therefore zero before the fault takes place.

⇒ ∆δ1 = ω1∆t = 56.25 × 0.05 = 2.8125 deg

δ1 = δ0 + ∆δ1 = 15.36 + 2.8125 = 18.1725 deg

KWCn v2.2 23


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Second interval:

Pa1= 1.0 − 0.0 = 1.0 pu

Or α1 =Pa1

M= 1.0 × 2250 = 2250 deg/sec2

⇒ ∆ω2 = α1∆t = 2250 × 0.05 = 112.5 deg/sec

ω2 = ω1 + ∆ω2 = 56.25 + 112.5 = 168.75 deg/sec

⇒ ∆δ2 = ω2∆t = 168.75 × 0.05 = 8.4375 deg

δ2 = δ1 + ∆δ2 = 18.1725 + 8.4375 = 26.61 deg

Successive intervals: Pa, α and ∆ω are the same as the second interval.

ω3 = ω2 + ∆ω3 = 281.25 deg/sec δ3 = 40.66 deg

ω4 = ω3 + ∆ω4 = 393.75 deg/sec δ4 = 60.34 deg

ω5 = ω4 + ∆ω5 = 506.25 deg/sec δ5 = 85.65 deg

ω6 = ω5 + ∆ω6 = 618.75 deg/sec δ6 = 116.58 deg

ω7 = ω6 + ∆ω7 = 731.25 deg/sec δ7 = 153.14 deg

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0

20

40

60

80

100

120

140

160

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35

Del

ta (

deg)

Time (s)

Rotor Angle

0.2582

Critical clearing angle= 90.62 deg

Hence, taken from the swing plot, the critical clearing time is 0.2582 sec.

KWCn v2.2 24


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3. Swing curve plots:

Once the circuit breaker is operated at the end of 0.2 s, the faulty section of the

system will be isolate and the system will operate on the power characteristic:

Pe = Pmax2sin δ = 2.716 sin δ

And the average accelerated power becomes :

Pa4=

Pa4−

+ Pa4+

2

=(1.0 − 0.0) + (1.0 − 2.716 sin 60.34)

2= −0.18 pu

Or α4 =Pa4

M= −0.18 × 2250 = −405 deg/sec2

⇒ ∆ω5 = α1∆t = −405 × 0.05 = −20.25 deg/sec

ω5 = ω4 + ∆ω5 = 393.75 − 20.25 = 373.5 deg/sec

⇒ ∆δ5 = ω5∆t = 373.5 × 0.05 = 18.67 deg

δ5 = δ4 + ∆δ5 = 60.34 + 18.67 = 79.01 deg

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Similarly, ω6 = 186.8 deg/sec δ6 = 88.34 deg

ω7 = 192.9 deg/sec δ7 = 88.04 deg

ω8 = −199.0 deg/sec δ8 = 78.10 deg

ω9 = −385.5 deg/sec δ9 = 58.82 deg

0

20

40

60

80

100

120

140

160

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

Del

ta (

deg)

Time (s)

Rotor Angle

KWCn v2.2 25


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5.5.2 Discontinuities due to Network Changes

A change in operating conditions causes a change in the value of Pa – hence

discontinuity occurs. Possible changes can be the removal of a fault or any line

switching operations. Depending on the exact ocurring moment of the discontinuity,

there are 3 possible cases.

1. The discontinuity occurs at the middle of an interval – this is simplest as no

special procedure is required.

timestep

nn-1n-2

Discontinuity

Pa(n)

Pa(n-1)Pa(n-2)

Pa

timestep

nn-1n-2

Discontinuity

Pa(n)

Pa(n-1)

Pa(n-2)

Pa

Case 1 Case 2

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2. The discontinuity occurs at the beginning of the interval – an average values of

the accelerating powers Pa(n)− and Pa(n)+ will be used instead, i.e.

Pa(n) = 12 (Pa(n)− + Pa(n)+)

where Pa(n)− and Pa(n)+ is the accelerating power immediately before and

after the change, e.g. clearing the fault.

3. The discontinuity occurs at some other than the beginning or the middle of an

interval – in this case, a weighted average value of Pa before and after the

discontinuity may be used.

In practice, a precise evaluation of Pa is not required in general as the time interval

used in calculation is so short that it is sufficiently accurate to consider the

discontinuity as either in case (1) or (2).

Alternatively, a variable time step approach can be used to adjust the the time

interval at which the discontinuity occurs such that the discontinuity locates exactly

at the middle of the interval.

KWCn v2.2 26


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5.6 Factors Affecting Transient Stability

• From the swing equation, the acceleration of the rotor is inversely proportional

to the inertia constant M of the machine when accelerating power is constant.

This means higher the inertia constant, the slower will be the change in the rotor

angle of the machine and thus large the critical clearing time. However, it is

uneconomical to improve the transient stability by increasing the inertia constant

and is normally not used.

• The methods normally used for improving the transient stability are:

1. Higher system voltage – an increase in system voltage results in higher

value of the steady-state stability limit (Pmax). The higher the Pmax value,

the smaller will be the transmission angle δ reqired to transfer a given

amount of power. This means the greater is the margin between the

steady-state transmission angle and the critical clearing angle.

2. Use of parallel lines and/or series capacitors to reduce the transmission

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reactance – again reducing the series reactance by using parallel lines

and/or series capacitors can achieve the same objective of improving the

Pmax.

3. Use of high speed circuit breakers and auto-reclosing breakers – the quicker

a breaker can operate, the faster the fault will be removed from the system

and the smaller will be the change in rotor position. Hence, the better the

tendency of the system to restore to normal operating conditions.

5.6.1 Automatic Voltage Regulator (AVR)

• A voltage regulator is the heart of the excitation system. The output voltage of

the generator changes only when the voltage regulator instructs the excitation

system to do so irrespective of the speed of response of the exciter.

• A regulator senses changes in the output voltage and/or current and causes

corrective action to take place. If the regulator is slow, the system will be a poor

one.

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• The settings and physical limits on the AVR will have a direct impact on the

system performance. With a good setting, both the steady-state and transient

stability limits can be improved with the use of AVR.

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5.6.2 Governor Control Effects on Stability

• Though speed governing control has little effects in terms of steady-steady

stability, fast acting governor can certainly improve multi-swing transient stabilty.

• In the short period ( 1s) after the disturbance, the governor and turbine will be

too slow to have any significant effect on the generator rotor response.

• However, the governing effects will kick in and improve the system response as

excess mechanical power coming the steam turbine has been reduced by the

closing the main steam controlling and interceptor valves.

KWCn v2.2 28

THE HONG KONG POLYTECHNIC UNIVERSITY Department of Electrical Engineering

EE4031 Power Systems Tutorial on Transient Stability

1. A three-phase fault is applied at P as shown in Fig.1 with reactances given on a common system base. The generator is delivering 1.0 pu power at the instant preceding the fault.

a) By calculation, find the critical clearing angle for clearing the fault with simultaneous opening of the breakers 1 and 2.

b) Plot the power-angle curves and show the equality of the accelerating and decelerating areas.

c) Suggest a method to improve the transient stability of this system and explain briefly how the method works.

Fig.1

2. A 50Hz generator with a reactance of 0.3 pu has an inertia constant H = 5. It delivers power of 0.8 pu to an infinite bus through a transmission network with transfer reactance of 0.2 pu. The internal voltage of the generator and the voltage of the infinite bus are maintained at 1.05 pu and 1 pu, respectively.

a) Determine the load angle of the generator.

b) Determine the steady-state stability limit.

c) Determine the synchronizing power coefficient.

d) Due to a fault on the transmission network, the transfer reactance is suddenly doubled. Plot the swing curve of the generator for the first 0.2 s after the fault using a time step of 0.05 s.

3. A generator operating at 50 Hz with an inertia constant H of 5 sec delivers half-load (i.e. Po = 0.5 pu) to an infinite bus of voltage 1.0 pu through a transmission link with total reactance XL of 0.4 pu. The internal voltage E and reactance XG of the generator is 1.5 pu and 0.8 pu, respectively.

a) Find the power angle δo and the steady-state stability limit Pmax.

b) A three-phase fault occurs at the generator terminal, find the critical clearing angle δcr and the corresponding critical clearing time tcr.

EE4031, KWCn, 23 Apr 2010 1

4. A generator with 0.2 pu reactance is connected to a busbar via a transformer having a reactance of 0.15 pu. The busbar is then connected to a large power system by a double-circuit line, each circuit of which has a reactance of 1.2 pu. The voltages at both the generator terminal and the large power system is 1.0 pu. The generator has an inertia constant _ of 5 sec and delivers 0.9 pu real power to the large power system.

a) Find the internal voltage and the power angle of the generator.

b) Due to malfunction of a relay, one of the circuits in the double circuit line was tripped suddenly and then reclosed after 0.2 sec. Calculate the power angle of the generator when the tripped circuit was reclosed, hence plot its value for the next 0.1 sec using time steps of 0.05 sec.

5. Fig.5 shows a synchronous generator connected to an infinite bus via a double circuit line with reactances and voltages given on a common system base. A transient three-phase fault occurs at the mid-point of one of the lines. At the instant preceding the fault, the generator is delivering a steady-state power Pg = 1.0 pu.

a) Determine the pre-fault steady-state rotor angle δo of the generator.

b) The fault is first isolated by switching out the faulty line when the rotor angle increases to δ1 = 90o, and is subsequently cleared by the time when the line is reclosed at rotor angle δ2 = 120o. Apply the equal area criterion, and determine the transient stability of the post-fault system.

Fig.5

6. A 50Hz generator with a reactance of 0.25 pu delivers a steady-state power of 0.8 pu

over a transmission system to an infinite busbar. The steady-state voltages at both the generator terminal and the infinite bus are 1.0 pu. The generator has an inertia constant H of 5 sec, and the effective transfer reactance of the transmission system is 0.4 pu.

a) Find the internal voltage and the power angle of the generator.

b) A solid three phase fault occurs in the transmission system and is subsequently cleared after 0.15 sec. The effective transfer reactances of the transmission system during and after the fault are 1.0 and 0.6 pu, respectively. Plot the power angle of the generator for the first 0.1 sec after the fault using time steps of 0.05 sec, and hence determine the power angle and power output of the generator after the fault has been cleared for 0.1 sec.

EE4031, KWCn, 23 Apr 2010 2

EE4031 Power Systems Tutorial on Transient Stability

1. a) The total transmission reactance:

0.2 0.4 // 0.4 0.05 0.45 pupreX = + + = 0.2 0.4 0.05 0.65 pupostX = + + =

The maximum power output:

max1.2 8 pu0.45 3pre

EVPX

= = = 2max

1.2 24 pu0.65 13post

EVPX

= = =

and their ratio

2max2

max

24 /13 98 / 3 13

Pr

P= = =

The initial rotor angle

1max

3sin 1 sin 0.3844 rad8o oP δ δ −= → = =

The maximum rotor angle

2

1max

13sin 1 sin 2.5692 rad24m mP δ δ −= → = =

The critical clearing angle

1 o2

2

( )sin coscos 1.2209 rad 69.95m o o mcr

rr

δ δ δ δδ − ⎡ ⎤− += =⎢ ⎥

⎣ ⎦=

b) power-angle curves

c) Install a “braking resistor” close to the generator terminals, the amount of energy available to accelerate the rotor can then be significantly reduced, and hence the system stability can be improved.

EE4031, KWCn, 23 Apr 2010 3

2. a) 0.3 0.2 0.5 pupreX = + =

max1.05 2.1 pu0.5pre

EVPX

= = =

-1 -1 omax

max

0.8sin sin sin 22.392.1

oo o o

PP PP

δ δ= → = = =

b) Stability limit max 2.1 puP= =

c) max cos 2.1cos22.39 1.94 pu/rados oP P δ= = =

d) 25 1 s /deg

180 180 50 1800HM

f= = =

× 0.3 0.4 0.7 pupostX = + =

(s)t maxEVPX

= max sineP P δ= 0.8a eP P= −

2

at P

M δ (deg)δ 0- 2.1 0.8 0 − − − 0+ 1.5 0.5714 0.2286 − − − 0avg − − 0.1143 0.51 0.51 22.90 0.05 1.5 0.5837 0.2163 0.97 1.48 24.38 0.10 1.5 0.6192 0.1808 0.81 2.29 26.67 0.15 1.5 0.6733 0.1267 0.57 2.86 29.53 0.20 1.5 0.7393 0.0607 0.27 3.13 32.66

where ( )

2

( ) ( 1) nn n at P

Mδ δ −= + and ( ) ( 1) ( )n n nδ δ δ−= +

22

24

26

28

30

32

34

0 0.05 0.1 0.15 0.2

Time (s)

Rot

or (d

eg)

EE4031, KWCn, 23 Apr 2010 4

3. a) The total transmission reactance: 0.8 0.4 1.2 puX = + =

The maximum power output:

max1.5 1.0 1.25 pu

1.2EVPX

×= = =

Steady-state output:

max sin 0.5 puo oP P δ= = ⇒ 1 0.5sin 23.58 0.4115 rad1.25oδ

−= = ° =

b) Fault-on:

⇒ 1max 0P = 1max

1max

0P

rP

= =

Post-fault:

⇒ 2max maxP P= 2max

2max

1P

rP

= =

The maximum rotor angle: 2.7301 radm oδ π δ= − =

The critical clearing angle:

1 o2

2

( )sin coscos 1.5599 rad 89.375m o o mcr

rr

δ δ δ δδ − ⎡ ⎤− += =⎢ ⎥

⎣ ⎦=

Considering the corresponding critical clearing time:

2 250 0.5 0.4115 1.5599 rad2 2 5

ocr cr o cr

f P t tH

π πδ δ× × × ×= × + = × + =

× ×

⇒ 0.3824 seccrt =

4. a) Let δt be the terminal voltage angle relative the infinite bus.

1 1 sin 0.9 42.4540.15 1.2 / 2 t tδ δ×

= ⇒ =+

°

Terminal current 1 42.454 1 0 0.9 0.350.75 0.75

t bV VI jj j− ∠ °− ∠ °

= = = + pu

Generator internal voltage: 1 (0.9 0.35)( 0.95) 0.668 0.855 1.085 52bE V IX j j j= + = + + = + = ∠ °

b) 25 1 / deg180 180 50 1800

HM sf

= = =×

1.20.2 0.15 0.952preX = + + = pu

0.2 0.15 1.2 1.55faultX = + + = pu post preX X=

EE4031, KWCn, 23 Apr 2010 5

t ∆t Pe (pu) Pa (pu) α (o/s2) ∆ω (o/s) ω (o/s) ∆δ (o) δ (o) 0– 0.9 0 0+ 0.5516 0.3484 0avg 0.05 0.1742 313.56 15.68 15.68 0.78 52.78 0.05 0.05 0.5575 0.3425 616.58 30.83 46.51 2.33 55.11 0.10 0.05 0.5742 0.3258 586.48 29.32 75.83 3.79 58.90 0.15 0.05 0.5994 0.3006 541.09 27.05 102.88 5.14 64.05 0.2– 0.6294 0.2706 0.2+ 1.0269 -0.1269 0.2avg 0.05 0.0718 129.32 6.47 109.35 5.47 69.51 0.25 0.05 1.0699 -0.1699 -305.77 -15.29 94.06 4.70 74.22 0.3 0.05 1.0990 -0.1990 -358.28 -17.91 76.15 3.81 78.02

where ( ) ( )( ) sinn n

e n

EVPX

δ= , ( ) ( ) ( )( )0.9 sinn n n

a m e n

EVP P PX

δ= − = − , ( )

( 1)n

n aPM

α + =

( 1) ( 1)n na tω α+ +Δ = Δ n, ( 1) ( ) ( 1)n nω ω ω+ += + Δ

( 1) ( 1)n n tδ ω+ +Δ = Δ n, ( 1) ( ) ( 1)n nδ δ δ+ += + Δ

5. a) Pre-fault transmission reactance 0 0.9 0.4 // ( 0.2 0.2) 1.1X j j j j j= + + =

Pre-fault power output 0 00

1.8 1sin sin 11.1

a bE VPX

δ δ0×

= = = ⇒ 0 37.67δ = °

b) Fault-on transmission reactance 10.9 0.4 0.4 0.2 0.2 0.9 3.1

0.2X j j× + × + ×

= =

⇒ 11

1.8 1sin sin 0.5806sin3.1

a bE VPX

δ δ δ×= = =

Fault-off transmission reactance 2 0.9 0.4 1.3X j j j= + =

⇒ 22

1.8 1sin sin 1.3846sin1.3

a bE VPX

×δ δ δ= = =

Post-fault transmission reactance = pre-fault transmission reactance = j1.1

⇒ 30

1.8 1sin sin 1.6364sin1.1

a bE VPX

δ δ δ×= = =

Total accelerating area

( )1

0

90

0 1 37.67( ) 1 0.5806sin 0.454aA P P d d

δ

δδ δ

°

°= − = − =∫ ∫ δ

=

Total decelerating area

( ) ( )

2 3

1 22 0 3 0

120 180 37.67

90 120

( ) ( )

1.3846sin 1 1.6364sin 1 0.256

dA P P d P P d

d d

δ δ

δ δδ δ

δ δ δ δ° °°−

° °

= − + −

= − + −

∫ ∫

∫ ∫

Based on the equal area criterion, the system is transient unstable as the allowable decelerating area is less the acceleration are, i.e. a dA A>

EE4031, KWCn, 23 Apr 2010 6

6. a) Let δt be the terminal voltage angle relative the infinite bus.

1 1sin 0.8 18.6630.4 t tδ δ×

= ⇒ = °

Terminal current 1 18.663 1 0 0.8 0.1310.4 0.4

jj j− ∠ ° − ∠ °

= = = +t bV VI pu

Generator internal voltage: 1 (0.8 0.131)( 0.25 0.4) 0.915 0.52 1.052 29.62j j j j= + = + + + = + = ∠bE V ΙX °

b) 25 1 / deg180 180 50 1800

HM sf

= = =×

0.25 0.4 0.65preX = + = pu

pu 0.250.25 1.0 1.25faultX = + = 0.6 0.85postX = + = pu

t ∆t Pe (pu) Pa (pu) α (o/s2) ∆ω (o/s) ω (o/s) ∆δ (o) δ (o) 0– 0.8 0 29.62 0+ 0.416 0.384 0avg 0.05 0.192 345.64 17.28 17.28 0.86 30.49 0.05 0.05 0.427 0.373 671.43 33.57 50.85 2.54 33.03 0.10 0.05 0.4587 0.3413 614.27 30.71 81.57 4.08 37.11 0.15– 0.5078 0.2922 0.15+ 0.7467 0.0533 0.15avg 0.05 0.1728 310.99 15.55 97.11 4.86 41.96 0.2 0.05 0.8276 -0.0276 -49.62 -2.48 94.63 4.73 46.70

where ( ) ( )( ) sinn n

e n

EVPX

δ= , ( ) ( ) ( )( )0.8 sinn n n

a m e n

EVP P PX

δ= − = − , ( )

( )n

n aPM

α =

, ( 1) ( )n na tω α+Δ = Δ n( 1) ( ) ( 1)n nω ω ω+ += + Δ

, ( 1) ( 1)n n tδ ω+ +Δ = Δ n( 1) ( ) ( 1)n nδ δ δ+ += + Δ

After fault cleared for 0.1 sec (i.e. t = 0.25 sec): Power angle = 46.7o

Power output = 0.9 pu

EE4031, KWCn, 23 Apr 2010 7

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